In Exercises 21–48, find the derivative of y with respect to t...

Question

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.

25. y=arcsec(2s+1)

Accepted Answer

Welcome back everyone. Find the derivative of y given that y equals inverse of 2 of 3 t minus 4. For this problem we want to find the derivative dyy divided by dt, right, and we have to notice that y is a composite function. So we will apply the chain rule. According to the chain rule, we take the derivative of the outer function inverse of sequence of 3 t minus 4 with respect to. The inner function, so with respect to 3 t minus 4 and we multiply by the derivative of the inner function 3 t minus 4 with respect to c. So now we are going to find each derivative. Let's begin with the derivative of inverse of 2, and let's remember that the derivative of inverse of sequent is going to be equal to. Now using the tables of derivatives, we will remember that it is equal to 1 divided by the absolute value of the inner function in this case 3 t minus 4. Multiplied by a square root of The square of the inner function, so that'd be 3 T minus 42 minus 1, right? And then we're multiplying by the derivative of 3 t minus 4. The derivative of -4 is 0. The derivative oft is 1, so we basically get 3, and overall we're multiplying by 3. So that would be 3 divided by. The absolute value of 3 t minus 4 multiplied by the square root of. 3 t minus 42 minus 1 and we're going to simplify slightly so what we can do is simply factor out 3 t minus 4 squared within the radical and this will be equal to 3 divided by The absolute value of 3 T minus 4. Multiplied by a square root of 3 t minus 42 we're going to have. One -1 divided by 3 t minus 42 Now that we have factored it out, we're going to use the properties of radicals to simplify further. So let's go ahead and extend the radical sign. And let's take the square root of 3t minus 42, which is the absolute value of 3t minus 4. So the final answer is going to be 3 divided by. The absolute value of 3 t minus 42 can be written as 3 t minus 42 without the absolute value, right, because it is a square of a number. And then we're going to have square root of. What we're left with is 1 minus. One divided by. 3 t minus 42 And that's going to be our final answer for this problem. Thank you for watching.

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.25. y=arcsec(2s+1)

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In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
25. y=arcsec(2s+1)